Question 12.12: Extrema of Rayleigh Line Consider the T-s diagram of Rayleig...
Extrema of Rayleigh Line
Consider the T-s diagram of Rayleigh flow, as shown in Fig. 12–50. Using the differential forms of the conservation equations and property relations, show that the Mach number is Ma_a = 1 at the point of maximum entropy (point a), and Ma _b=1 \sqrt{k} at the point of maximum temperature (point b).
It is to be shown that Ma_a = 1 at the point of maximum entropy and Ma _b=1 \sqrt{k} at the point of maximum temperature on the Rayleigh line.
Assumptions The assumptions associated with Rayleigh flow (i.e., steady onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects) are valid.
Analysis The differential forms of the continuity (𝜌V = constant), momentum [rearranged as P + (𝜌V )V = constant], ideal gas (P = 𝜌RT ), and enthalpy change (Δh = c_p ΔT ) equations are expressed as
\rho V=\text { constant } \rightarrow \rho d V+V d \rho=0 \rightarrow \frac{d \rho}{\rho}=-\frac{d V}{V} (1)
P+(\rho V) V=\text { constant } \rightarrow d P+(\rho V) d V=0 \rightarrow \frac{d P}{d V}=-\rho V (2)
P=\rho R T \rightarrow d P=\rho R d T+R T d \rho \rightarrow \frac{d P}{P}=\frac{d T}{T}+\frac{d \rho}{\rho} (3)
The differential form of the entropy change relation (Eq. 12–40) of an ideal gas with constant specific heats is
s_2-s_1=c_P \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1} (12.40)
d s=c_p \frac{d T}{T}-R \frac{d P}{P} (4)
Substituting Eq. 3 into Eq. 4 gives
d s=c_p \frac{d T}{T}-R\left(\frac{d T}{T}+\frac{d \rho}{\rho}\right)=\left(c_p-R\right) \frac{d T}{T}-R \frac{d \rho}{\rho}=\frac{R}{k – 1} \frac{d T}{T}-R \frac{d \rho}{\rho} (5)
since
c_p-R=c_v \quad \rightarrow \quad k c_v-R=c_v \quad \rightarrow \quad c_v=R /(k-1)
Dividing both sides of Eq. 5 by dT and combining with Eq. 1,
\frac{d s}{d T}=\frac{R}{T(k – 1)}+\frac{R}{V} \frac{d V}{d T} (6)
Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rearranging,
\frac{d T}{d V}=\frac{T}{V}-\frac{V}{R} (7)
Substituting Eq. 7 into Eq. 6 and rearranging,
\frac{d s}{d T}=\frac{R}{T(k – 1)}+\frac{R}{T – V^2 / R}=\frac{R\left(k R T – V^2\right)}{T(k – 1)\left(R T – V^2\right)} (8)
Setting ds/dT = 0 and solving the resulting equation R(kRT − V²) = 0 for V give the velocity at point a to be
V_a=\sqrt{k R T_a} \quad \text { and } \quad Ma _a=\frac{V_a}{c_a}=\frac{\sqrt{k R T_a}}{\sqrt{k R T_a}}=1 (9)
Therefore, sonic conditions exist at point a, and thus the Mach number is 1.
Setting dT /ds = (ds /dT )^{−1} = 0 and solving the resulting equation T (k − 1) × (RT − V^2) = 0 for velocity at point b give
V_b=\sqrt{R T_b} \quad \text { and } \quad Ma _b=\frac{V_b}{c_b}=\frac{\sqrt{R T_b}}{\sqrt{k R T_b}}=\frac{1}{\sqrt{k}} (10)
Therefore, the Mach number at point b is Ma _b=1 \sqrt{k} . For air, k = 1.4 and thus Ma_b = 0.845.
Discussion Note that in Rayleigh flow, sonic conditions are reached as the entropy reaches its maximum value, and maximum temperature occurs during subsonic flow.